IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

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A UNIT SQUARE IS DRAWN WITH A(0,0) B(1,0) C(1,1) D(0,1)...A QUADRILATERAL HAS BEEN DRAWN WITH THE VERTICES ON THE SIDES OF SQUARE AND THE SIDES OF IT ARE a,b,c,d as shown in the figure...

prove that

a^2 + b^2 + c^2 + d^2 is greater than or equal to 2 and less than or equal to 4...

pink_ele

look,
a^2 + b^2 + c^2 + d^2=CQ^2+CP^2+PB^2+BS^2+AS^2+AR^2+RC^2+CQ^2<=
AB^2+BC^2+CD^2+AD^2
THIS GIVES
a^2 + b^2 + c^2 + d^2 =<4

allamraju

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Let AS=x,BP=y,CQ=z,DR=t then

$\sum$ a²= $\sum$ x²+(1-x)²= $\sum$ 2x²-2x+1=(1/2) $\sum$ (2x-1)²+1 $\ge$ (1/2) $\sum$ 1=2

Also, $\sum$ 2x²-2x+1= $\sum$ 1-2x(1-x) $\le$ $\sum$ 1=4

Hence,2 $\le$ $\sum$ a²4

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Let AS=x,BP=y,CQ=z,DR=t then

a2=x2+(1-x)2=2x2-2x+1=(1/2)(2x-1)2+1(1/2)1=2

Also,2x2-2x+1=1-2x(1-x)1=4

Hence,2a2 4

$\sum$ a²= $\sum$ x²+(1-x)²= $\sum$ 2x²-2x+1=(1/2) $\sum$ (2x-1)²+1 $\ge$ (1/2) $\sum$ 1=2

Also, $\sum$ 2x²-2x+1= $\sum$ 1-2x(1-x) $\le$ $\sum$ 1=4

Hence,2 $\le$ $\sum$ a²4